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What is the square footage of area inside a regulation 400m track?
Wiki User
∙ 12y ago
Fun question.
IAAF standards require 84.39 straightaways on both sides and 115.61m curves, measured from either 20 cm or 30 cm from the inside of the first lane, depending upon whether there is a no curb or rail or if there is a curb or rail. This should not change the area of the space inside the point of measurement, though.
The lengths of the two curves add to a full circle, so 2 x pi x radius = 231.22. Solving for r gets you 36.8, which means the width of the internal rectangle is 73.6 (2 x radius).
The area of the internal rectangle is 84.39 x 73.4 (length x width), with a product of 6211.1 (approximate, since pi was used in the calculations).
The area of the two halves of a circle on each end is pi x 36.8 squared (pi x radius squared), with a product of 4254.47 meters squared (again, approximate since pi was used in the calculations)
The sum of the two areas is 10,465.57 meters squared (approx.)
Wiki User
∙ 12y ago
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What is the square footage inside a 200 meter running track?
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It takes ten identical pieces to form a circular track for a pair of toy racing cars If the inside of each piece is 3.4 inches shorter than the outside arc what is the width of the track?
Length of outer track = sum of length of 10 pieces = circumference of the outer circleif R is the Radius of outer circle then...Circumference of the outer track = 2pi*RSimilarly the circumference of the inner track (with radius r) = 2pi*rlength of each outer piece is 3.4 inch more than length of inner pieceSo total outer length is 10*3.4 =34 inches more than the inner length.=> Outer Circumference - Inner Circumference = 34 inches=> 2pi*R - 2pi*r = 34=> 2pi(R -r) = 34=> R-r = 34/2pi = 5.41 inches=> R-r = Width of the track = 5.41 inches1 year ago
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75000
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